Wow, Nice Problem! I claim that $L = 1$. Firstly, We need $L$ to be \[ \sum_{i = 1}^{\infty} \frac{1}{a_i} \]for obvious reasons. Now, I just need to prove that \[ \sum_{i = 1}^{\infty} \frac{1}{a_i} = 1 \]Notice that $a_{n + 1} - 1 = a_n^2 - a_n = a_n(a_n - 1)$ \[ \sum_{i = 1}^{\infty} \frac{1}{a_i - 1} - \frac{1}{a_{i}} = \sum_{i = 1}^{\infty} \frac{1}{a_i (a_i - 1)} = \sum_{i = 1}^{\infty} \frac{1}{a_{i + 1} - 1} = \sum_{i = 2}^{\infty} \frac{1}{a_i - 1} \]Therefore, \[ \sum_{i = 1}^{\infty} \frac{1}{a_i} = \frac{1}{a_1 - 1} = 1 \]

I would like to present the following intuitive solution i came up with . Hope i didn't make any silly mistake . $\underline{Solution:}$ We claim that $L=1$. For this consider the following: Claim 1: $\Pi_{i=1}^j \ {a_i}= a_{j+1}-1$ Proof : By the given equation, $a_{k + 1} - 1 = a_k^2 - a_k = a_k(a_k - 1)=a_k(a_{k-1}^2-a_{k-1})=a_ka_{k-1}(a_{k-1}-1)=.....=\Pi_{i=1}^k a_i \ \ \blacksquare$ . Claim 2: The final value of the summation for each $k$ is the of the form $$\frac{\Pi_{i=1}^ka_i -1}{\Pi_{i=1}^ka_i}$$where $k \ \epsilon \ N$. Proof: We induct on $k$. Note that for $k=1,2$ , the sums are $\frac{1}{2}$ and $\frac{5}{6}$, respectively, completing the base cases. Now for the induction part, assume it is true for some $k=g$, Then \begin{align} \sum_{i=1}^{g+1} \frac{1}{a_i} =& \frac{\Pi_{i=1}^ga_i -1}{\Pi_{i=1}^ga_i} + \frac{1}{a_{g+1}} \\ =& \frac{\Pi_{i=1}^{g+1}a_i- a_{g+1}+\Pi_{i=1}^{g}a_i}{\Pi_{i=1}^{g+1}a_i} \\ =& \frac{\Pi_{i=1}^{g+1}a_i -1}{\Pi_{i=1}^{g+1}a_i} \end{align} Where the last step follows from the first claim. This completes the induction $\blacksquare$. From these claims it clearly follows that each term is just less than 1, therefore we are done, $QED$. Motivation : the first thing i did was to evaluate the first few terms and the respective sums, and the pattern ( which is proved in claim 2) immediately popped up as each number being of the form $\frac{b-1}{b}$ for some $b$. A second later, one obviously realises that b must be the product of the terms since two consecutive numbers have $gcd=1$. Then i tried to induct and realised immediately that i would need ("what is now") claim 1, whose proof took another three seconds to realise since taking the 1 from RHS to LHS in the given equation gives us exactly the desired form. Honestly though, posting the solution took thrice as long as solving it !! Edit: Could someone please rectify the error in this latex command ? I don't know whats wrong, seems to working perfectly on maths stack exchange as well as the demo section in mathjax.

Psyduck909 wrote: Edit: Could someone please rectify the error in this latex command ? I don't know whats wrong, seems to working perfectly on maths stack exchange as well as the demo section in mathjax. On AoPS you should not put dollar signs around the `align` environment, I have fixed this for you.

v_Enhance wrote: Psyduck909 wrote: Edit: Could someone please rectify the error in this latex command ? I don't know whats wrong, seems to working perfectly on maths stack exchange as well as the demo section in mathjax. On AoPS you should not put dollar signs around the `align` environment, I have fixed this for you. Ahh understood, Thanks a lot Evan.

Here is mine solution just for storage My Solution: Claim: Lets claim that $L=1$ Proof: so we know that i have to prove is \[\sum_{i=1}^\infty \frac{1}{a_i}=1\]we know that $a_{n+1}=a_{n}^2-a_{n}+1\implies a_{n+1}-1=a_n(a_n-1)$ so we can see in the sequence like\[\implies \frac{1}{a_1-1}+\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots=\frac{-1}{a_1(a_1-1)}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\dots\]\[\implies \frac{-1}{a_2-1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\dots=\frac{-1}{a_2(a_2-1)}+\frac{1}{a_3}+\frac{1}{a_4}+\frac{1}{a_5}+\dots\]so basically by seeing this we can do is \[\sum_{i=1}^\infty\frac{1}{a_i-1}-\frac{1}{a_i}=\sum_{i=1}^\infty\frac{1}{a_{i}(a_i-1)}=\sum_{i=1}^\infty\frac{1}{a_{i+1}-1}=\sum_{i=2}^\infty\frac{1}{a_i-1}\]so we can clearly see that \[\sum_{i=1}^\infty \frac{1}{a_i}=1\]hence $L=1$ which means the following inequality is less than $1$.$\blacksquare$

Cheeky problem. I don't even know why I pulled out such dense analysis details on this one now. I claim $s_k=\sum_{t=1}^k\frac{1}{a_t}=1-\frac{1}{a_k^2-a_k}$, and then the rest basically falls into place. I’ll do the analysis bookwork at the end to finish the problem. However this is almost trivial by pulling out induction; note $s_1=\frac{1}{2}=1-\frac{1}{a_1^2-a_1}$, and $s_{k+1}=s_k+\frac{1}{a_{k+1}}=1-\frac{1}{a_k^2-a_k}+\frac{1}{a_k^2-a_k+1}=1-\frac{1}{(a_k^2-a_k)(a_k^2-a_k+1)}=1-\frac{1}{a_{k+1}(a_{k+1}-1)}=1-\frac{1}{a_{k+1}^2-a_{k+1}}$. Now we pull out the analysis bookwork. Note $a_{k+1}\ge2>0$; this is by induction. In fact I claim $a_k$ is an increasing sequence, but note that $a_{k+1}-a_k=(a_k-1)^2\ge0$, done. Since $a_{1}=2>0$, we have our positivity claim. Now this proves that $s_k$ is increasing. But $0<s_k<1$ for all $k$, the positivity by induction and $s_1>0$, the second as $s_k=1-\frac{1}{a_k(a_k-1)}<1$ since $a_k,a_k-1>0$. Hence $s_k$ is a bounded and monotonically increasing, so by the monotonic sequence theorem, $\lim_{k\to\infty}s_k$ has a limit. Call it $L\le1$. Now clearly then $\frac{1}{a_k-1}$ has a limit (this is by rearranging; we get this since $1-\frac{1}{a_k(a_k-1)}=1-\frac{1}{a_{k+1}-1}$ has a limit), which can be seen to be $1-L$. However, notice $\frac{1}{a_k-1}$ is then a strictly decreasing sequence (again by rearranging), so $\frac{1}{a_k-1}\ge1-L$, or $a_k\le\frac{1}{1-L}+1=\frac{2-L}{1-L}$. If $L<1$ then $a_k$ is bounded from above, which cannot be true. This is because $a_k$ is a strictly increasing sequence (by the above) of positive integers. Now it cannot be the case we have a strictly increasing, positive, bounded integer sequence since otherwise by repeated applications of the discrete inequality $a_k-a_1\ge k-1$, thereby establishing that $a_k$ has no upper bound (if we have upper bound $U$, choose $k>U+1-a_1$ to get a contradiction). Hence $L\ge1$. But $L\le1$, so $L=1$. Therefore $\lim_{k\to\infty}s_k=1$. Hence we have that for all $L'\ge1$, $s_k\le L'$ for all $k\ge1$ by strictly increasing. Now by the fact that $(s_k)$ is a Cauchy sequence we have that for all $\epsilon>0$ there exist $N>0$ such that $1-s_k=\left|s_k-L\right|<\epsilon$ for all $k>N$. Hence $s_k>1-\epsilon$ for all $k>N$. If $L'<1$, let $\epsilon=1-L'$, and apply the previous insight to win. Therefore $L=1$. aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Of course I could have just chickened out and do a summation manipulation, but this can’t always be done in more practical examples.